Ask question

Ask question

All categories

Nookie1986 [14]

2 years ago

5

The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=4900

0 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical. (a) What is the magnitude of the lift force? (b) Determine the magnitude of the air resistance R that opposes the motion.

1 answer:

Sedaia [141]2 years ago

3 0

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

Explanation:

Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then

(a) in the vertical direction we have

$L\cos(20\deg)-W=0\\L=\frac{W}{\cos(20\deg)}=\frac{49000 N}{\cos(20\deg)}\approx \mathbf{52144.71 N}$ .

(b) Now horizontally,

$L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.$

You might be interested in

You lower the temperature of a sample of liquid carbon disulfide from 90.3 ∘ C until its volume contracts by 0.507 % of its init

Lady_Fox [76]

Answer:

$T_{f}$ = 85.89 ° C

Explanation:

The linear thermal expansion process is given by

ΔL = L α ΔT

For the three-dimensional case, the expression takes the form

ΔV = V β ΔT

Let's apply this equation to our case

ΔV / V = -0.507% = -0.507 10-2

ΔT = (ΔV / V) 1 /β

ΔT = -0.507 10⁻² 1 / 1.15 10⁻³

ΔT = -4.409

$T_{f}$ –T₀ = 4,409

$T_{f}$ = T₀ - 4,409

$T_{f}$ = 90.3-4409

$T_{f}$ = 85.89 ° C

6 0

2 years ago

Read 2 more answers

A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the

cupoosta [38]

Answer: deceleration of $1.904\ rad/s^2$

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. $a=\alpha \cdot r$

Using the equation of motion

$v^2-u^2=2as\\$

Insert $v=0,u=20,s=300$

$0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2$

Write acceleration as $a=\alpha \cdot r$

$-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2$

So, the car must be experiencing the deceleration of $1.904\ rad/s^2$ .

4 0

2 years ago

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of you

olchik [2.2K]

PART A)

horizontal distance that will be moved = 14 m

Height of the fence = 5.0 m

height from which it is thrown = 1.60 m

angle of projection = 54 degree

So here we can say that stone will travel vertically up by distance

$\Delta y = 5 - 1.6 = 3.40 m$

now we will have displacement in horizontal direction

$\Delta x = 14 m$

now we know that

$v_x = vcos54$

$v_y = vsin54$

now we will have

$\Delta x = v_x t$

$14 = (vcos54)t$

also for y direction

$\Delta y = v_y t + \frac{1}{2}at^2$

$3.40 = (vsin54)t - \frac{1}{2}(9.8) t^2$

now from the two equations we will have

$3.40 = (vsin54)(\frac{14}{vcos54}) - 4.9 t^2$

$3.40 = 14 tan54 - 4.9 t^2$

$3.40 = 19.3 - 4.9 t^2$

$t = 1.8 s$

now from above equations

$14 = vcos54 (1.8)$

$v = 13.2 m/s$

So the minimum speed will be 13.2 m/s

Part B)

Total time of the motion after which it will land on the ground will be "t"

so its vertical displacement will be

$\Delta y = -1.60 m$

now we will have

$-1.60 = v_y t + \frac{1}{2}at^2$

$-1.60 = (13.2sin54)t - \frac{1}{2}(9.8)t^2$

$4.9 t^2 - 10.7t - 1.60 = 0$

$t = 2.3 s$

Now the time after which it will reach the fence will be t1 = 1.8 s

so total time after which it will fall on other side of fence

$t_2 = t - t_1$

$t_2 = 2.3 - 1.8 = 0.5 s$

now the displacement on the other side is given as

$\Delta x = (vcos54) t_2$

$\Delta x = (13.2 cos54)(0.5)$

$\Delta x = 3.88 m$

4 0

2 years ago

A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

$s=1.63 m$

5 0

2 years ago

ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on it

Nadya [2.5K]

Answer:

K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

Em₀ = K = ½ mv²

final point. When it is at tea = 50º

Em_f = K + U

Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

h = L - L cos 50

Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

Em₀ = Em_f

½ mv² = ½ m v_b² + mg L (1 - cos50)

K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

K_b = 36 +42.0

K_b = 78 J

4 0

2 years ago